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When you can't find a single positive value in the pivot column it means that the linear program is unbounded. In other words, you can increase the objective function indefinitely while maintaining a feasible solution. So, if this happens, your algorithm shouldn't backtrack and try to find another candidate for the entering variable; it should tell the program to stop and declare that the problem is unbounded. If you are getting this situation on a problem you know is bounded then there is a bug somewhere else in your code.

Let's look at an example to see why no positive values in the pivot column means the LP is unbounded.

Letting $x_2$ enter the basis means increasing $x_2$. Normally, we increase the entering variable until doing so any more would make one of the current basic variables negative (and thus the current solution infeasible). However, the fact that there are no positive values in $x_2$'s column means that increasing $x_2$ does not cause any of the current basic variables to decrease: the value of $x_3$ remains at $1$, the value of $x_5$ actually increases with $x_2$, and the values of $x_1$ and $x_4$ remain $0$. Thus we can increase $x_2$ indefinitely without any other variable going negative, which means we can increase the objective function indefinitely in this direction without the current solution ever becoming infeasible. This means the problem is unbounded.